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Question Number 113811 by Aina Samuel Temidayo last updated on 15/Sep/20

If log_(12) 27=a, express log_6 16 in terms of a.

$$\mathrm{If}\:\mathrm{log}_{\mathrm{12}} \mathrm{27}=\mathrm{a},\:\mathrm{express}\:\mathrm{log}_{\mathrm{6}} \mathrm{16}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\mathrm{a}. \\ $$

Answered by bemath last updated on 15/Sep/20

((ln 27)/(ln 12)) = a ⇒((3ln 3)/a) = ln 12 ((3ln 3)/a) = ln 3+2ln 2 → 2ln 2=((3ln 3)/a)−ln 3 ⇔ ((ln 16)/(ln 6)) = ((4ln 2)/(ln 2+ln 3)) = (((2(3−a)ln 3)/a)/((((3−a)ln 3)/(2a))+ln 3)) = (((6−2a)/a)/((3−a+2a)/(2a))) = ((6−2a)/a) ×((2a)/(3+a)) = ((12−4a)/(3+a))

$$\frac{\mathrm{ln}\:\mathrm{27}}{\mathrm{ln}\:\mathrm{12}}\:=\:{a}\:\Rightarrow\frac{\mathrm{3ln}\:\mathrm{3}}{{a}}\:=\:\mathrm{ln}\:\mathrm{12} \\ $$$$\frac{\mathrm{3ln}\:\mathrm{3}}{{a}}\:=\:\mathrm{ln}\:\mathrm{3}+\mathrm{2ln}\:\mathrm{2}\:\rightarrow\:\mathrm{2ln}\:\mathrm{2}=\frac{\mathrm{3ln}\:\mathrm{3}}{{a}}−\mathrm{ln}\:\mathrm{3} \\ $$$$\Leftrightarrow\:\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{ln}\:\mathrm{6}}\:=\:\frac{\mathrm{4ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}}\:=\:\frac{\frac{\mathrm{2}\left(\mathrm{3}−{a}\right)\mathrm{ln}\:\mathrm{3}}{{a}}}{\frac{\left(\mathrm{3}−{a}\right)\mathrm{ln}\:\mathrm{3}}{\mathrm{2}{a}}+\mathrm{ln}\:\mathrm{3}} \\ $$$$=\:\frac{\frac{\mathrm{6}−\mathrm{2}{a}}{{a}}}{\frac{\mathrm{3}−{a}+\mathrm{2}{a}}{\mathrm{2}{a}}}\:=\:\frac{\mathrm{6}−\mathrm{2}{a}}{{a}}\:×\frac{\mathrm{2}{a}}{\mathrm{3}+{a}}\:=\:\frac{\mathrm{12}−\mathrm{4}{a}}{\mathrm{3}+{a}} \\ $$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

=((4(3−a))/(3+a)) Thanks.

$$=\frac{\mathrm{4}\left(\mathrm{3}−\mathrm{a}\right)}{\mathrm{3}+\mathrm{a}} \\ $$$$ \\ $$$$\mathrm{Thanks}. \\ $$

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